Question

这是我的pojo课

inner class Data : Serializable {
    var abc: ArrayList<ABC>? = null
    var xyz: ArrayList<XYZ>? = null

    inner class XYZ : Serializable {
        var x: String? = null
        var date: String? = null
    }

    inner class ABC : Serializable {
        var a: String? = null
        var date: String? = null
    }
}

我想将ABC和XYZ数组列表合并到一个列表中，并根据日期对该列表进行排序

到目前为止，我已经完成了

val items = ArrayList<Any>()
items.addAll(data!!.xyz!!)   
items.addAll(data!!.abc!!)

并获得这样的合并列表

x (from XYZ arraylist in descending order)
x (from XYZ arraylist in descending order)
x (from XYZ arraylist in descending order)
a (from ABC arraylist in descending order)
a (from ABC arraylist in descending order)
a (from ABC arraylist in descending order)

但是我想要的是根据日期对整个列表进行排序

Answer 1

最好是引入一个WithDate接口，让您的类实现它。

interface WithDate {
    val date: String?
}

inner class XYZ : Serializable, WithDate {
    var x: String? = null
    override var date: String? = null
}

现在您有了WithDate的列表，可以使用sortBy函数了。

val list = ArrayList<WithDate>()
...
list.sortBy(WithDate::date)

如果您不能更改类，请执行以下操作：

val list = ArrayList<Any>()
list.sortBy {
    when(it) {
        is Data.XYZ -> it.date ?: ""
        is Data.ABC -> it.date ?: ""
        else -> ""
    }
}

Answer 2

您必须使用Collections.sort对列表进行排序。步骤1：合并列表时，请使用

   Collections.sort(<UR LIST>, new compareMethod());

步骤2：创建compareMethod（）;

class compareMethod implements Comparator<LISTNAME> {
    @Override
    public int compare(LISTNAME object1, LISTNAME object2) {
        return object1.getDate().compareTo(object2.getDate());
    }
}

希望您能理解

Answer 3

请尝试以下方法。希望它能解决您的问题。

首先参加一个像下面这样的普通pojo课

class MergedResponse : Serializable {

var data: Data? = null

inner class Data : Serializable {

@SerializedName(value = "xyz", alternate = arrayOf("abc"))
var mCommon: ArrayList<Common>? = null

inner class Common : Serializable {
        var x: String? = null
        var a: String? = null
        var date: String? = null
        }
    }
}

然后在您的课程中编写以下代码

val mainData = Gson().toJson(data!!)
var mJsonXYZ: JSONObject = JSONObject(mainData)
var mJsonABC: JSONObject = JSONObject(mainData)
mJsonXYZ.remove("abc")
mJsonABC.remove("xyz")

val mypojoXYZ = Gson().fromJson<MergedResponse.Data>(Gson().toJson(Gson().fromJson<Data>(mJsonXYZ.toString(), Data::class.java)), MergedResponse.Data::class.java)

val mypojoABC = Gson().fromJson<MergedResponse.Data>(Gson().toJson(Gson().fromJson<Data>(mJsonABC.toString(), Data::class.java)), MergedResponse.Data::class.java)

val itemsXYZ = ArrayList<MergedResponse.Data.Common>()
itemsXYZ.addAll(mypojoXYZ.mCommon!!)
itemsXYZ.addAll(mypojoABC.mCommon!!)


Collections.sort(itemsXYZ, object : Comparator<MergedResponse.Data.Common> {
    override fun compare(object1: MergedResponse.Data.Common, object2: MergedResponse.Data.Common): Int {
    return object1.date!!.compareTo(object2.date!!)
}
})

根据日期将排序列表与2种不同的Pojo合并

3 个答案: